Q. 40 - Calculus

Question

For each pair of functions in Exercises 40–45, (a) Algebraically find all values of $\theta$ where$f_1\left(\theta \right)=f_2\left(\theta \right)$  (b) Sketch the two curves in the same polar coordinate system. (c) Find all points of intersection between the two curves. 40. $f_1\left(\theta \right) = \sin 2\theta and f_2\left(\theta \right) = -\sin 2\theta .$

Step-by-Step Solution

Answer

Part (a)$\theta =\frac{n\mathrm{\pi }}{2}$

Part (c) $\left(\pm \frac{\mathrm{\pi }}{2},0\right) ; \left(\pm \mathrm{\pi },0\right)...$

Part (b)

Part (a) Step 1. Given Information.

The two curves are :

$f_1\left(\theta \right) = \sin 2\theta and f_2\left(\theta \right) = -\sin 2\theta .$

Part (a) Step 2. Algebraically.

When

 $$\begin{gathered} f_1\left(\theta \right)=f_2\left(\theta \right) \\ \sin \left(2\theta \right)=-\sin \left(2\theta \right) \\ \sin \left(2\theta \right) + \sin \left(2\theta \right) =0 \\ 2\sin \left(2\theta \right)=0 \\ \sin \left(2\theta \right) =0 \\ \sin \theta =\sin 0 = n\pi \\ 2\theta =n\pi \\ \theta =\frac{n\mathrm{\pi }}{2}. \end{gathered}$$

Part (b) Step 1. Sketching the curves.

Consider the given information from part (a).

The two curves are :

Part (c) Step 1. Point of intersection.

Consider the given information from part (a).

The points of intersection between the curves are 

$$\begin{gathered} f_1\left(\theta \right)=f_2\left(\theta \right) \\ \sin \left(2\theta \right)=-\sin \left(2\theta \right) \\ \sin \left(2\theta \right) + \sin \left(2\theta \right) =0 \\ 2\sin \left(2\theta \right)=0 \\ \sin \left(2\theta \right) =0 \\ \sin \theta =\sin 0 = n\pi \\ 2\theta =n\pi \\ \theta =\frac{n\mathrm{\pi }}{2}. \end{gathered}$$

So, Points are $\left(\pm \frac{\mathrm{\pi }}{2},0\right) ; \left(\pm \mathrm{\pi },0\right),..$